The area bounded by the curves y = -sqrt{-x} | vedclass

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(B) Given curves are $y = -\sqrt{-x}$ and $x = -\sqrt{-y}$ for $x, y \le 0$.Squaring both sides,we get $y^2 = -x$ and $x^2 = -y$.These are parabolas o

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is $1/3$

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(B) Given curves are $y = -\sqrt{-x}$ and $x = -\sqrt{-y}$ for $x, y \le 0$.Squaring both sides,we get $y^2 = -x$ and $x^2 = -y$.These are parabolas opening in the third quadrant.To find the intersection points,substitute $y = -x^2$ into $y^2 = -x$:$(-x^2)^2 = -x \Rightarrow x^4 = -x \Rightarrow x(x^3 + 1) = 0$.Since $x \le 0$,the intersection points are $x = 0$ and $x = -1$.When $x = 0, y = 0$. When $x = -1, y = -1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 0$:$A = \int_{-1}^{0} (	ext{upper curve} - 	ext{lower curve}) dx$In the interval $[-1, 0]$,the curve $y = -\sqrt{-x}$ is above $y = -x^2$.$A = \int_{-1}^{0} (-\sqrt{-x} - (-x^2)) dx = \int_{-1}^{0} (x^2 - \sqrt{-x}) dx$$A = [\frac{x^3}{3} - \frac{(-x)^{3/2}}{3/2} 	imes (-1)]_{-1}^{0} = [\frac{x^3}{3} + \frac{2}{3}(-x)^{3/2}]_{-1}^{0}$$A = (0 + 0) - (\frac{-1}{3} + \frac{2}{3}(1)) = -(-\frac{1}{3} + \frac{2}{3}) = -\frac{1}{3}$.Since area must be positive,we take the absolute value: $A = 1/3$.

The area bounded by the curves $y = -\sqrt{-x}$ and $x = -\sqrt{-y}$ where $x, y \le 0$ is:

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